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3 years ago

What process is occurring in the picture?

Physics

1 answer:

Alex777 [14]3 years ago

3 0

Two fat black arrows are swimming along together, when they see a single skinny black arrow coming toward them. They are afraid of strangers, and they know that the skinny one must be mean and tough if it's not afraid to travel alone. So they turn to the side and get out of the skinny arrow's way.

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Ilia_Sergeevich [38]

Below is the attachment of the solutions.

5 0

3 years ago

I need help understanding this concept . Would appreciate it so much. Thank you

Firlakuza [10]

Answer:

Explanation:

Both these questions are based on the Universal Law of Gravitation, which is given by :

F = Gm1m2 / r²

2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5

F = 6.67 x 10⁻⁵ x 8 / 1.5

F = 35.57 x 10⁻⁵ N

3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴

F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29

F = 63.13 x 10⁻³ N

7 0

2 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

A student is preparing to bake a cake at home and places several ingredients on the countertop to use. First he preheats the ove

Evgen [1.6K]

Answer:

baking the cake batter

Explanation:

Baking the cake batter will indicate that chemical change has occurred here. What is a chemical change?

A chemical change is one in which a new kind of matter is formed.
It is usually accompanied by energy either evolution or absorption of energy in form of heat or light or both.
The process is irreversible.
When the batter bakes, a new substance different from the cake mix is obtainable.
We cannot get back the ingredient from this baked cake. It is impossible.
This is good indicator of chemical change.

7 0

3 years ago

If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet

Evgesh-ka [11]

Answer:

$S = \dfrac{1}{2.5}$